「学习总结-Haskell-1」Haskell 基础知识

Haskell基础知识

1 Haskell 基础知识

1.1 Haskell 开发环境安装

1.2 Haskell 代码如何运行

  • 代码
-- helloWorld.hs
-- 定义名为hi的函数
hi = "hello, world!"
  • 运行
$ ghci
GHCi, version 7.4.1: http://www.haskell.org/ghc/  :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
ghci>:l helloWorld.hs
[1 of 1] Compiling Main             ( helloWorld.hs, interpreted )
Ok, modules loaded: Main.
ghci>hi
"hello, world!"

1.3 Haskell 函数定义和使用

-- function.hs
doubleMe x = x + x
doubleUs x y = doubleMe x + doubleMe y

ghci>:l function.hs 
[1 of 1] Compiling Main             ( function.hs, interpreted )
Ok, modules loaded: Main.
ghci>doubleMe 2
4
ghci>doubleUs 3 4
14

1.4 Haskell 基本语句和编程模式

  • if语句
maxNumber x y = if x > y then x
                else y

ghci>:l baseStruc.hs
[1 of 1] Compiling Main             ( baseStruc.hs, interpreted )
Ok, modules loaded: Main.
ghci>maxNumber 3 5
5
ghci>maxNumber 5 3
5
  • case语句
isLucky x = case x of 1 -> "1:lucky"
                      2 -> "2:No lucky"
                      otherwise -> "lucky"

ghci>isLucky 1
"1:lucky"
ghci>isLucky 2
"2:No lucky"
ghci>isLucky 3
"lucky"
ghci>isLucky 4
"lucky"
  • let语句
doubleSum x y =
  let doublex = 2 * x
      doubley = 2 * y
  in doublex + doubley

ghci>doubleSum 3 4
14
ghci>doubleSum 1 2
6
  • where语句
doubleSum2 x y = doublex + doubley
  where doublex = 2 * x
        doubley = 2 * y

ghci>doubleSum2 1 2
6
ghci>doubleSum2 2 3
10
  • 模式匹配(Pattern Matching)

例1

lucky 7 = "You are lucky"
lucky x = "Sorry, you are not lucky"

ghci>lucky 7
"You are lucky"
ghci>lucky 1
"Sorry, you are not lucky"
ghci>lucky 2
"Sorry, you are not lucky"

例2

foo 0 x = (-1) * x
foo 1 x = x
foo x y = 0

ghci>foo 1 2 
2
ghci>foo 2 3
0
ghci>foo 3 4
0
ghci>foo 0 2
-2

例3

foo' 0 x = (-1) * x
foo' x y = 0
foo' 1 x = x

ghci>foo' 1 2
0
ghci>foo' 1 3
0
ghci>foo' 2 3
0
ghci>foo' 3 4
0
ghci>foo' 0 2
-2
  • 门卫(Guard)

例1

mdivide x y
  | y == 0 = "Can not divide"
  | x / y > 10 = "first number is larger than the second number"
  | x / y < 1  = "first number is less than the second number"
  | otherwise = "almost equal"

ghci>mdivide 3 0
"Can not divide"
ghci>mdivide 3 2
"almost equal"
ghci>mdivide 30 2
"first number is larger than the second number"
ghci>mdivide 3 5
"first number is less than the second number"

例2

mdivide' x y
  | x / y > 10 = "first number is larger than the second number"
  | y == 0 = "Can not divide"
  | x / y < 1  = "first number is less than the second number"
  | otherwise = "almost equal"

ghci>mdivide' 3 0
"first number is larger than the second number"
ghci>mdivide' 3 1
"almost equal"
ghci>mdivide' 30 3
"almost equal"
ghci>mdivide' 30 2
"first number is larger than the second number"
ghci>mdivide' 2 3
"first number is less than the second number"
  • 递归(Recursion)

注：理解门卫与递归如何结合在一起工作

msum x y
  | x > y = msum y x
  | x == y = x
  | otherwise = x + msum (x+1) y

ghci>msum 1 100
5050
ghci>msum 2 4
9
ghci>msum 100 1
5050

posted @ 2013-01-31 23:05  Iambda  阅读(336)  评论(0)  编辑  收藏  举报